Question

If $\Delta ABC$ and $\Delta PQR$ are two similar triangles shown in the figure. AM and PN are the medians on $\Delta ABC$ and $\Delta PQR$respectively. The ratio of areas of $\Delta ABC$ and $\Delta PQR$ is 9:25. If AM = PO = 5 cm. Find the value of 3(ON) in cm ___

Answer 1

We are given two triangles i.e. $\Delta ABC$ and $\Delta PQR$ such that $\Delta ABC\sim \Delta PQR$ so, the corresponding ratio of sides and corresponding angles should be equal.
$\Rightarrow$ $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$
Also, $\frac{\frac{1}{2}\times BC}{\frac{1}{2}\times QR}=\frac{BM}{QN}$
Since both the triangles i.e. $\Delta ABC$ and $\Delta PQR$ are similar, there angles will be equal i.e. $\angle$ A = $\angle$ P, $\angle$ B = $\angle$ Q and $\angle$ C = $\angle$ R.
In $\Delta ABM$ and $\Delta PQN$,
$\frac{AB}{PQ}=\frac{BM}{QN}$
$\angle$ B = $\angle$ Q
Therefore $\Delta ABM\sim \Delta PQN$ [ SAS similarity]
Given AM = PO = 5 cm
$\Rightarrow$ $\frac{Areaof\Delta ABC}{Areaof\Delta PQR}=\frac{A{B}^2}{P{Q}^2}=\frac{9}{25}\Rightarrow \frac{AB}{PQ}=\frac{3}{5}$
$\Rightarrow$ $\frac{AB}{PQ}=\frac{AM}{PN}=\frac{3}{5}\Rightarrow \frac{5}{5+ON}=\frac{3}{5}$
Then
$25=15+3\left(ON\right)ON=\frac{10}{3}3\left(ON\right)=10cm$
Hence the length of 3(ON) is 10 cm.